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Thread: Convergence in L1 and characteristic function

  1. #1
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    Convergence in L1 and characteristic function

    If $\displaystyle \mu (E_n) < \infty $ for $\displaystyle n \in \mathbb {N} $, and $\displaystyle 1_{E_n} \rightarrow f $ in $\displaystyle L^1 $, then f equals to the characteristic function of a measurable set.

    Proof so far.

    So I know that $\displaystyle \int \mid 1_{E_n} - f \mid d \mu \rightarrow 0 $

    And I need to find a measurable set, say M, such that $\displaystyle f=1_M$

    I apologize for the lack of significant amount of work here, but I just don't really know how to start here, any hints?

    Thank you.
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  2. #2
    Senior Member Shanks's Avatar
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    Actually , I think , f equals to the characteristic function of a measurable set, a. e.
    Hint :convergence in $\displaystyle L^1$ implies convergence in measure, and convergence in measure implies a subsequence convergence a. e.
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